Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 4

Year : 2013

Let D (a) be the given integral then by the rule of differentiation under the integral sign.

$\dfrac {dI}{da}=\int\limits_0^{\infty}=\dfrac {\partial b}{\partial a}=\int\limits_0^{\infty}\dfrac 1{x^2}.\dfrac 1{11+ax^2}x^2 dx$

$\int\limits_0^{\infty}\dfrac 1{11+ax^2} dx$

$\dfrac 1a\int\limits_0^{\infty}\dfrac 1{\dfrac{11}a+x^2} $

$=\dfrac 1a. \sqrt{\dfrac a{11}}\Bigg[\tan^{-1} \dfrac x{\sqrt{\dfrac {11}a}}\Bigg]^\infty_0$

$=\Bigg[\dfrac 1{a\times 11}\times \dfrac \pi2\Bigg]$

$=\dfrac \pi2 \times \dfrac 1{\sqrt{11a}}$